Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=\cos ^{3} x-1.25 \cos ^{2} x+0.225\). Find \(f^{\prime}\left(x_{0}\right)\) at that point \(x_{0}\) in \([\pi / 2, \pi]\) where \(f\left(x_{0}\right)=0\).

Step-by-Step Solution

Answer

At the point \(x_0 = \frac{2\pi}{3}\), \(f'(x_0) = -\sqrt{3}\).

Step 1: Finding the Root

We need to find the value of \(x_0\) in the interval \([\pi/2, \pi]\) where \(f(x_0) = 0\). Given \(f(x) = \cos^3 x - 1.25 \cos^2 x + 0.225\), set \(f(x)\) to zero: \(\cos^3 x - 1.25 \cos^2 x + 0.225 = 0\).

Step 2: Solving for \(\cos x\)

To solve the equation \(\cos^3 x - 1.25 \cos^2 x + 0.225 = 0\), let's denote \(y = \cos x\). The equation becomes \(y^3 - 1.25y^2 + 0.225 = 0\). We need to find \(y\) for \(x\) in \([\pi/2, \pi]\). This interval corresponds to \(y \in [-1, 0]\).

Step 3: Finding Real Roots

Evaluate the polynomial \(y^3 - 1.25y^2 + 0.225 = 0\) for \(y\) in \([-1,0]\). One potential approach is testing small rational numbers. It turns out \(y = -0.5\) is a root (it satisfies the equation when plugged in). Thus, \(x_0\) could be when \(\cos x_0 = -0.5\).

Step 4: Determining \(x_0\)

With \(\cos x_0 = -0.5\) and \(x_0\) in \([\pi/2, \pi]\), \(x_0\) must be \(\frac{2\pi}{3}\) since \(\cos(\frac{2\pi}{3}) = -0.5\).

Step 5: Differentiate \(f(x)\)

Differentiate \(f(x) = \cos^3 x - 1.25 \cos^2 x + 0.225\) using the chain rule. The derivative of \(\cos^3 x\) is \(3\cos^2 x(-\sin x)\), and the derivative of \(-1.25 \cos^2 x\) is \(-2.5 \cos x(-\sin x)\). Combine these derivatives: \[f'(x) = -3\cos^2 x \sin x + 2.5 \cos x \sin x\].

Step 6: Simplifying the Derivative

Factor \(\sin x\) from the derivative: \[f'(x) = (-3\cos^2 x + 2.5\cos x) \sin x\].

Step 7: Evaluate \(f'(x_0)\)

Substitute \(\cos x_0 = -0.5\) and \(x_0 = \frac{2\pi}{3}\) into the derivative:- Calculate \(f'(x_0) = (-3(-0.5)^2 + 2.5(-0.5))\sin \left(\frac{2\pi}{3}\right)\).- \(-3(-0.5)^2 = -0.75\), \(2.5(-0.5) = -1.25\), and \(\sin \left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\).- Thus, \(f'(x_0) = (-0.75 - 1.25) \frac{\sqrt{3}}{2} = -2 \cdot \frac{\sqrt{3}}{2} = -\sqrt{3}\).

Key Concepts

Roots of Trigonometric Functions

Understanding the roots of trigonometric functions involves finding the values of the variable that make the entire expression equal to zero. In our particular exercise, we dealt with a function involving the cosine trigonometric identity: \( f(x) = \cos^3 x - 1.25 \cos^2 x + 0.225 \).
To find the root, we first set \( f(x_0) = 0 \) and solved for \( x_0 \) within the interval \([\pi/2, \pi]\).
The interval \([\pi/2, \pi]\) is important because it helps us narrow down the possible values of \( \cos x \), which range between -1 to 0 in this interval.
Eventually, we discovered that \( x_0 = \frac{2\pi}{3} \) because for this value, \( \cos x_0 = -0.5 \), satisfying the equation \( \cos^3 x - 1.25 \cos^2 x + 0.225 = 0 \).
Recognizing these points where the function equals zero is essential for various calculus applications, including finding derivatives at specific points.

Chain Rule

The chain rule is a vital tool in calculus used to differentiate composite functions. When a function is composed of an outer function and an inner function, the derivative of the overall function can be determined using the chain rule.
In practice, this means if you have a function like \( f(x) = \cos^3 x \), you can consider the outer function as something cubed and the inner function as \( \cos x \).
The derivative \( f(x) \) using the chain rule is calculated by differentiating the outer function, multiplying by the derivative of the inner function. This involves:

• Taking the derivative of \( u^3 \), where \( u = \cos x \), which is \( 3u^2 \)
• Multiplying by the derivative of \( \cos x \), which is \( -\sin x \)

Thus, we get \( -3 \cos^2 x \sin x \).
This approach can extend to multiple composite functions and lets us break down more complex differentiation tasks efficiently.

Polynomial Equation Solving

Polynomial equation solving is a fundamental skill in differential calculus that allows you to break down and find the solutions to polynomial equations. Our exercise involved resolving a polynomial in terms of \( y = \cos x \), specifically \( y^3 - 1.25y^2 + 0.225 = 0 \).
The solution process involves identifying potential real roots, often by using methods like trial and error for rational numbers, which in this case led us to discover \( y = -0.5 \) fits within the interval \([-1, 0]\).
With \( \cos x_0 = -0.5 \), we then reverse-engineer this back to finding the corresponding \( x \)-value, \( x_0 = \frac{2\pi}{3} \) in the given interval. This process allows us to identify specific values where the function equals zero, which are key for subsequent derivative calculation and analysis.

Derivative Evaluation

Derivative evaluation involves calculating the rate at which a function changes at a specific point—an essential concept in differential calculus. In our example, after finding where \( f(x_0) = 0 \), we needed \( f'(x_0) \).
This required differentiating our original function \( f(x) = \cos^3 x - 1.25 \cos^2 x + 0.225 \) using previously discussed techniques like the chain rule.
Once the derivative was simplified to \( f'(x) = (-3\cos^2 x + 2.5\cos x) \sin x \), we substituted the already calculated root \( \cos x_0 = -0.5 \).

• The expressions \( -3(-0.5)^2 \) and \( 2.5(-0.5) \) simplify to \(-0.75 \) and \( -1.25 \) respectively.
• Applying these values to the simplified derivative expression and factoring in \( \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} \), we evaluated the derivative to be \(-\sqrt{3}\).

Evaluating derivatives precisely allows us to understand function behavior at critical points and can inform decisions about slope and concavity on graphs.